Related Fixed Point Theorems in Partially Ordered b -Metric Spaces and Applications to Integral Equations

In this research paper, we have set some related ﬁ xed point results for generalized weakly contractive mappings de ﬁ ned in partially ordered complete b -metric spaces. Our results are an extension of previous authors who have already worked on ﬁ xed point theory in b -metric spaces. We state some examples and one sample of the application of the obtained results in integral equations, which support our results.


Introduction
The concept of b-metric space has been dealt with by distinctly different authors since it first appeared and became largely used. Bakhtin in [1] was the first who introduced this concept, and later on, Czerwik in [2,3] invested it in the convergence of measurable functions. Accordingly, several interesting results about the existence of fixed points have been obtained concerning both single-valued and multivalued operators in b-metric spaces (see, e.g., [4,5]). In the same way, Hussain and Shah [6] conducted a research and then came up with results from KKM mappings in cone b-metric spaces. Likewise, Roshan et al. [7] used the notion of almost generalized contractive mappings in ordered complete b-metric spaces and set some fixed and common fixed point results. Regarding partially ordered metric spaces, Ran and Reurings (see [8]) had first set up their assumptions before Nieto and Rodríguez-López used them (see [9,10]). Afterwards, many authors presented several interesting and significant results in such spaces (see [4,7,[11][12][13][14][15][16][17][18][19]).
The purpose of this research paper is to prove some fixed point theorems for generalized contractive conditions for four mappings in complete b-metric spaces.
Our work goes through the following steps. First, we have demonstrated a two b-metric space theorem with four mappings. Second, we have stated a relevant example which backs up the mentioned theorem. Third, we have given three corollaries related to the theorem. Besides, an example with only two related corollaries is provided for the second theorem. To sum up, we conclude the manuscript by an application to solve a system of integral equations.
Throughout this paper, ℝ and ℝ + denote the sets of all real numbers and nonnegative real numbers, respectively. Consistent with [3,4], the following definitions and outcomes are going to be vital and required in the ending.
Definition 2 (see [20]). Let ðX, dÞ be a b-metric space. Then, a sequence fx n g in X is (a) convergent if there exists x ∈ X such that dðx n , xÞ ⟶ 0 as n ⟶ ∞. In this case, we write lim n⟶∞ x n = x.
Definition 3. Let X be a nonempty set. Then, ðX, d, δ,≼Þ is called a partially ordered two b-metric space if d and δ are b-metrics on the partially ordered set ðX, ≼Þ.
A subset Y of a partially ordered set X is said to be well ordered if every two elements of Y are comparable.
Definition 4 (see [4]). Let ðX, ≼Þ be a partially ordered set. A mapping f on X is called dominating (resp. dominated) if x≼f x (resp. f x≼x) for each x in X.
Definition 5. Let fx n g be a sequence in a b-metric space ðX, dÞ, g, h : X ⟶ X, and x ∈ X.
(i) x is said to be a coincidence point of pair fg, hg if gx = hx (ii) fg, hg is said to be compatible if dðghx n , hgx n Þ ⟶ 0 as n ⟶ ∞ (iii) fg, hg is said to be weakly compatible if ghx = hgx, where gx = hx Remark 6 (see [20]). In a b-metric space ðX, dÞ, the following assertions hold: (R1) A convergent sequence has a unique limit (R2) Each convergent sequence is a Cauchy sequence (R3) In general, a b-metric is not continuous (R4) In general, a b-metric does not induce a topology on X.
The fact in (R3) requires the following Lemma concerning b-convergent sequences to prove our findings: Lemma 7 (see [4]). Let ðX, dÞ be a b-metric space with s ∈ ½1 ,∞Þ and suppose that fx n g and fy n g are convergent to x, y, respectively. Then, we have In particular, if x = y, then we have lim n⟶∞ dðx n , y n Þ = 0.
Moreover, for each z ∈ X, we have

Main Results
Throughout this paper, let f , g, P, and T be four self-maps on X, and d and δ are two b-metrics with constant s and r, respectively. Set Finally, we consider the following assumptions: Assumption 8. Let fx n g and fy n g be two sequences such that (i) fx n g is nonincreasing (ii) ½for all n, y n ≼x n and y n → u ⇒ ½u≼x n , for all n.
Assumption 9. Let f , g, P, and T be four self-maps on X such that either condition (a) fg, Tg is weakly compatible, f f , Pg is compatible and f or P is continuous, or (b) f f , Pg is weakly compatible, fg, Tg is compatible and g or T is continuous.
Our main result is the following theorem: Theorem 10. We consider four self-mappings f , g, P, and T in ordered complete two b-metric space ðX, d, δ, ≼Þ that fulfill the following conditions: (i) ff , gg is dominated and fP, Tg is dominating (ii) f X ⊆ TX and gX ⊆ PX (iii) ψ ∈ Ψ, φ ∈ Φ and for every two comparable elements x, y ∈ X, we have (iv) Assumptions 8 and 9 are satisfied Then, f , g, P, and T have a unique common fixed point in X.
Proof. Let x 0 be an arbitrary point in X. By condition (ii), we can define inductively two sequences fx n g and fy n g in X as follows: We have Thus, x n+1 ≼x n for all n ≥ 0, and Step 1. If y 2k+1 = y 2k for some k, then ðy n Þ n is constant. Indeed, we have which implies If we insert in (10), we obtain φðdðy 2k+1 , y 2k+2 ÞÞ = 0, and then fy n g is a constant sequence. Its value is a common fixed point of f , g, P, and T: In the following, we can assume that dðy 2n , y 2n+1 Þ > 0 for each n.
Assume that there exist ε, τ > 0 for which we can find subsequences fy 2m k g and fy 2n k g of fy 2n g such that 2n k > 2 m k > k for all k, we have Using the triangle inequality in b-metric spaces, we obtain which leads to and since we obtain By the same arguments, we obtain and consequently Similarly, we obtain Abstract and Applied Analysis Moreover, we have Using (24) and (26)-(28), we obtain By the same arguments, we obtain Using (31), we get Then, Similarly, we obtain Consequently, which implies that It follows that

Abstract and Applied Analysis
We have also Then, Step 4. y is a common fixed point of f , g, P, and T.
On the other hand, we have f X ⊆ TX, then there exists a point v ∈ X such that f y = Tv: Abstract and Applied Analysis Moreover, v≼Tv = f y≼y gives If n ⟶ +∞, and using Lemma (7), we obtain If n ⟶ +∞ in (56), and using Lemma (7), we obtain which implies dðy, gyÞ = δðy, gyÞ = 0 and then y = gy.
We conclude that f y = gy = Py = Ty = y. If f is continuous, the proof is the same. The same goes for condition (b) of Assumption 9.
Step 5. Suppose that u and v are two common fixed points of f , g, P, and T but dðu, vÞ > 0 with u and v are comparable. We have Hence, ψðdðu, vÞÞ ≤ ψðdðu, vÞÞ − φðdðu, vÞÞ: Therefore, u = v.
Example 11. Let X = ½0, 1 ; we define on X the b-metrics dðx, yÞ = ðx − yÞ 2 with s = 2 and δðx, yÞ = jx − yj 3 with r = 4 . We endow X with the partial order ≼ given by And we define f , g, P, and T on X by Obviously, conditions (i) and (ii) of Theorem 10 are satisfied. Moreover, ff , Pg is weakly compatible, fg, Tg is compatible and g is continuous.
If we take P and T as the identity maps on X in Theorem (10), we conclude the following corollary.

Corollary 12.
We consider two self-mappings f and g in ordered complete two b-metric space ðX, d, δ,≼Þ that fulfill the following conditions: (i) f f , gg is dominated (ii) ψ ∈ Ψ, φ ∈ Φ and for every two comparable elements x, y ∈ X, we have (iii) Assumption 8 is satisfied Then f and g have a unique common fixed point in X.
If we take ψðtÞ = t for t ∈ ½0,∞Þ in Corollary (12), we have the following corollary.

Corollary 13.
We consider two self-mappings f and g in ordered complete two b-metric space ðX, d, δ,≼Þ that fulfill the following conditions: (i) f f , gg is dominated (ii) φ ∈ Φ and for every two comparable elements x, y ∈ X, we have (iii) Assumption 8 is satisfied Then, f and g have a unique common fixed point in X.

Corollary 14.
We consider two self-mappings f and g in ordered complete two b-metric space ðX, d, δ,≼Þ that fulfill the following conditions: (i) f f , gg is dominated (ii) k ∈ ð0, 1Þ and for every two comparable elements x, y ∈ X, we have Theorem 15. We consider two self-mappings f , g, P, and T in ordered complete two b-metric space ðX, d, δ,≼Þ that fulfill the following conditions: (i) f f , gg is dominated and fP, Tg is dominating (ii) f X ⊆ TX and gX ⊆ PX 8 Abstract and Applied Analysis (iii) ψ ∈ Ψ, φ ∈ Φ and for every two comparable elements x, y ∈ X, we have (iv) Assumptions 8 and 9 are satisfied Then, f , g, P, and T have a unique common fixed point in X.
Proof. If we follow similar arguments to those given in Theorem (10), we have the following steps.
Step 1. If y 2k+1 = y 2k for some k, then ðy n Þ n is constant.
In the following, we can assume that dðy 2n , y 2n+1 Þ > 0 for each n.
Assume that there exists ε > 0 for which we can find subsequences fy 2m k g and fy 2n k g of fy 2n g such that n k is the smallest index for which 2n k > 2m k > k, dðy 2m k , y 2n k Þ ≥ ε, and dðy 2m k −2 , y 2n k Þ < ε.
Following the same argument as in Theorem (10) As + a 5 d f x 2n k , Tx 2m k −1 À Á = a 1 d y 2n k , y 2m k −1 + a 2 d y 2n k +1 , y 2n k + a 3 d y 2m k , y 2m k −1 + a 4 d y 2n k , y 2m k + a 5 d y 2n k +1 , y 2m k −1 : As Taking the upper limit as k ⟶ ∞ and using (85) and (89), we obtain ψ εs 3 À Á ≤ ψ s 4 lim sup k⟶∞ d y 2n k +1 , y 2m k ≤ ψ lim sup which implies that So, lim inf k⟶∞ Q d ðx 2n k , x 2m k −1 Þ = 0, which contradicts (90). It follows that fy 2n g is a Cauchy sequence in X. Since X is complete, there exists y ∈ X so that Step 4. y is a common fixed point of f , g, P, and T.
Since P is continuous, we have Using the triangle inequality in b-metric space, we get